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E
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Difficulty:
45%
(medium)
Question Stats:
76% (03:20) correct
24%
(03:17)
wrong
based on 591
sessions
History
Date
Time
Result
Not Attempted Yet
Maths, Physics and chemistry books are stored on a library shelf that can accommodate 25 books. Currently, 20% of the shelf spots remain empty. There are twice as many maths books as physics books and the number of physics books is 4 greater than that of chemistry books. Among all the books, 12 books are soft cover and the remaining are hard-cover. If there are a total of 7 hard-cover books among the maths and physics books. What is the probability, that a book selected at random is either a hard cover book or a chemistry book?
A. 1/10
B. 3/20
C. 1/5
D. 1/4
E. 9/20
A. 1/10
B. 3/20
C. 1/5
D. 1/4
E. 9/20
We know that M = 2P
P = 4 + C
M = 8+2C
8 + 2c + 4 + c + c = 20
4c + 12 = 20
4c = 8
c = 2 , P = 6, M = 12
Please assist further. Thanks
P = 4 + C
M = 8+2C
8 + 2c + 4 + c + c = 20
4c + 12 = 20
4c = 8
c = 2 , P = 6, M = 12
Please assist further. Thanks
30 Mar 2014, 10:32
Kudos
Bookmarks
First phase of this problem requires you to determine how many mathematics and chemistry books are even on the shelf. To do so, you have the equations:
m + p + c = 20 (since 4/5 of the 25 spots are full of books)
m = 2p
p = 4 + c
From that, you can use Substitution to get everything down to one variable.
c = p - 4
m = 2p
p = p
Then (p - 4) + 2p + p = 20, so 4p = 24 and p = 6. That means that there are 12 math, 6 physics, and 2 chemistry books on the shelf.
With those numbers, you also know that there are 8 total hardcovers, 1 of which is chemistry. So if your goal is to get either a hardcover or a chemistry, there are 9 ways to "win" - either one of the 7 hardcovers that aren't chemistry or the two chemistry books. So out of the 20 total, 9 provide the desired outcome, making the answer E.
Note - a common trap answer here is A, as people forget that "2 chem + 8 hard" double counts the one book that is "both". Venn Diagram logic can be helpful to avoid that trap, or in cases like this with relatively small numbers it may be even more convenient to just write out the possibilities.
m + p + c = 20 (since 4/5 of the 25 spots are full of books)
m = 2p
p = 4 + c
From that, you can use Substitution to get everything down to one variable.
c = p - 4
m = 2p
p = p
Then (p - 4) + 2p + p = 20, so 4p = 24 and p = 6. That means that there are 12 math, 6 physics, and 2 chemistry books on the shelf.
With those numbers, you also know that there are 8 total hardcovers, 1 of which is chemistry. So if your goal is to get either a hardcover or a chemistry, there are 9 ways to "win" - either one of the 7 hardcovers that aren't chemistry or the two chemistry books. So out of the 20 total, 9 provide the desired outcome, making the answer E.
Note - a common trap answer here is A, as people forget that "2 chem + 8 hard" double counts the one book that is "both". Venn Diagram logic can be helpful to avoid that trap, or in cases like this with relatively small numbers it may be even more convenient to just write out the possibilities.
30 Mar 2014, 19:44
Kudos
Bookmarks
Mackieman
Everything is correct except this: P(AnB) = P(A)*(B) = (1/10)*(2/5) = 2/50
P(AnB) is given by the product of individual probabilities when the events are independent.
How many Chemistry books are hardcover? There are total 8 hardcovers and 7 belong to Math and Physics. This means there is only 1 Chemistry hardcover book out of total 20 books.
So P(AnB) = 1/20
Now P(AuB) = 1/10 + 2/5 - 1/20 = 9/20
